一位德明政府中学校友计划捐资在正信园中央建造两个锦鲤池。这两个水池的形状分别为“D”和“H”，如下图所示（所有尺寸单位均为米，图示未按比例绘制）。 阴影区域表示水池表面。“D”形池由一个 $3h$ 乘 $h$ 的矩形和由两个半径分别为 $2r$ 和 $r$ 的同心半圆形成的弯曲区域组成。“H”形池由七个边长为 $h$ 的相同正方形组成。 证明两个水池的总表面积 $A\text{ m}^2$ 由下式给出： $A = 10h^2 + \tfrac{3}{2}\pi r^2.$ 景观设计师建议 $h$ 和 $r$ 满足方程 $rh^2 = k$，其中 $k$ 是常数。随着 $r$ 和 $h$ 的变化，求出 $A$ 取得驻值时 $\dfrac{r}{h}$ 的精确值。确定该驻点的性质，并在 $k = 1$ 时利用计算器求出 $A$ 的驻值。 曲线 $C$ 的方程由 $y = 2x + 5 + \dfrac{4}{x}$ 给出。 通过求导，求出使曲线 $C$ 上没有点对应的 $y$ 的精确值集合。

1. Find the total surface area of the rectangular and square components of the ponds.

$\text{Area of H pond} = 7 \times h^2 = 7h^2$
$\text{Area of rectangular part of D pond} = 3h \times h = 3h^2$
$\text{Total straight-edged area} = 7h^2 + 3h^2 = 10h^2$

2. Find the area of the curved component of the D pond (the semi-circular annulus).

$\text{Area of outer semi-circle} = \frac{1}{2}\pi(2r)^2 = 2\pi r^2$
$\text{Area of inner semi-circle} = \frac{1}{2}\pi(r)^2$
$\text{Area of annulus} = 2\pi r^2 - \frac{1}{2}\pi r^2 = \frac{3}{2}\pi r^2$

3. Sum the areas to obtain the final expression for the total surface area $A$.

$A = 10h^2 + \frac{3}{2}\pi r^2$ (shown)

1. Express the total area $A$ in terms of a single variable, $r$.

Given $rh^2 = k$, we have $h^2 = \frac{k}{r}$.
Substitute this into the expression for $A$ from part (i):
$A = 10h^2 + \frac{3}{2}\pi r^2$
$A = 10\left(\frac{k}{r}\right) + \frac{3}{2}\pi r^2$
$A = 10k r^{-1} + \frac{3}{2}\pi r^2$

2. Differentiate $A$ with respect to $r$ and set it to zero for a stationary point.

$\frac{\mathrm{d}A}{\mathrm{d}r} = -10k r^{-2} + 3\pi r$
For a stationary value, $\frac{\mathrm{d}A}{\mathrm{d}r} = 0$:
$-\frac{10k}{r^2} + 3\pi r = 0$
$3\pi r = \frac{10k}{r^2}$
$3\pi r^3 = 10k$

3. Substitute $k = rh^2$ back to find the exact value of the ratio $\frac{r}{h}$.

$3\pi r^3 = 10(rh^2)$
Since $r \neq 0$, divide both sides by $r$:
$3\pi r^2 = 10h^2$
$\frac{r^2}{h^2} = \frac{10}{3\pi}$
Since $r$ and $h$ represent physical lengths, they must be positive. Thus:
$\frac{r}{h} = \sqrt{\frac{10}{3\pi}}$

4. Determine the nature of the stationary point using the second derivative test.

$\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = 20k r^{-3} + 3\pi = \frac{20k}{r^3} + 3\pi$
Since $k$ is a positive constant and $r > 0$, we have $\frac{20k}{r^3} > 0$.
Therefore, $\frac{\mathrm{d}^2A}{\mathrm{d}r^2} > 0$, which means the stationary point is a minimum.

5. Evaluate the stationary value of $A$ when $k = 1$.

When $k = 1$, the stationary condition is:
$3\pi r^3 = 10(1) \implies r = \left(\frac{10}{3\pi}\right)^{\frac{1}{3}} \approx 1.0199$
Substitute $r$ and $k=1$ into the expression for $A$:
$A = \frac{10(1)}{r} + \frac{3}{2}\pi r^2$
$A = \frac{10}{1.0199} + \frac{3}{2}\pi (1.0199)^2$
$A \approx 9.8048 + 4.9018 = 14.7066$
$A = 14.7 \text{ m}^2 \text{ (to 3 s.f.)}$

1. Differentiate the equation of the curve with respect to $x$.

$y = 2x + 5 + 4x^{-1}$
$\frac{dy}{dx} = 2 - 4x^{-2} = 2 - \frac{4}{x^2}$

2. Set the first derivative to zero to find the $x$-coordinates of the stationary points.

$2 - \frac{4}{x^2} = 0$
$x^2 = 2 \implies x = \sqrt{2} \text{ or } x = -\sqrt{2}$

3. Calculate the corresponding $y$-coordinates and determine the nature of each stationary point using the second derivative test.

$\frac{d^2y}{dx^2} = \frac{8}{x^3}$
For $x = \sqrt{2}$, $y = 2\sqrt{2} + 5 + \frac{4}{\sqrt{2}} = 5 + 4\sqrt{2}$.
Since $\frac{d^2y}{dx^2} = \frac{8}{(\sqrt{2})^3} > 0$, $( \sqrt{2}, 5 + 4\sqrt{2} )$ is a local minimum.

For $x = -\sqrt{2}$, $y = -2\sqrt{2} + 5 + \frac{4}{-\sqrt{2}} = 5 - 4\sqrt{2}$.
Since $\frac{d^2y}{dx^2} = \frac{8}{(-\sqrt{2})^3} < 0$, $( -\sqrt{2}, 5 - 4\sqrt{2} )$ is a local maximum.

4. Determine the set of values of $y$ for which there are no points on the curve.

Given that the local maximum is at $y = 5 - 4\sqrt{2}$ and the local minimum is at $y = 5 + 4\sqrt{2}$, the graph takes all values except those strictly between these two turning points.

Thus, there are no points on $C$ for:
$5 - 4\sqrt{2} < y < 5 + 4\sqrt{2}$